Occupation games on graphs in which the second player takes almost all vertices

Given a connected graph G=(V,E), two players take turns occupying vertices v∈V by placing black and white tokens so that the current vertex sets B,W⊆V are disjoint, B∩W=0?, and the corresponding induced subgraphs G[B] and G[W] are connected any time. A player must pass whenever (s)he has no legal move. (Obviously, after this, the opponent will take all remaining vertices, since G is assumed connected.) The game is over when all vertices are taken, V=B^∗∪W^∗. Then, Black and White get b=|B^∗|/|V| and w=|W^∗|/|V|, respectively. Thus, the occupation game is one-sum, b+w=1, and we could easily reduce it to a zero-sum game by simply shifting the payoffs, b^′=b−1/2,w^′=w−1/2. Let us also notice that b≥0 and w≥0; moreover, b>0 and w>0 whenever |V|>1.[Let us remark that the so-called Chinese rules define similar payoffs for the classic game of GO, yet, the legal moves are defined in GO differently.]Like in GO, we assume that Black begins. It is easy to construct graphs in which Black can take almost all vertices, more precisely, for each ε>0 there is a graph G for which b>1−ε. In this paper we show that, somewhat surprisingly, there are also graphs in which White can take almost all vertices.